Do discrete valuation rings correspond to local rings of points in fibre? Given projective curves $C$ and $C'$ and a surjective morphism $\varphi\colon C\to C'$, such that $Q\in C'$ is a smooth point and its fibre $\varphi^{-1}(Q)$ consists of smooth points.
Then $\mathcal{O}_{C',Q}$ and $\mathcal{O}_{C,P}$, for $\varphi(P) = Q$, are discrete valuation rings.
My question is: Are all discrete valuation rings in $k(C)$ containing $\mathcal{O}_{C',Q}$ of the form $\mathcal{O}_{C,P}$, for some $P\in\varphi^{-1}(Q)$?
If this is not true under this conditions, can one make additional assumptions, such that this is true? And a counterexample, if my claim is wrong, would be nice.
I already asked this on math.stackexchange, but didn't get any answer.
 A: There are DVRs between $k[x]_{(x)}$ and $k[[x]]$ that are transcendental over $k(x)$.  For example, the $x$-adic valuation DVR of $k(x,y)$ with $y=\Sigma x^{n!}$ or $y=e^x$ (characteristic 0).
A: Answering my own question is weird to me, but still, I have come to this point (and still need some help):
Lemma 1 Let $F\subset F'$ be function fields (i.e. finite algebraic extensions of $k(X)$) over an algebraically closed field $k$, such that $F'/F$ is a (finite) algebraic extension. Let $(A,\mathfrak{m}_A)$ in $F$ and $(B,\mathfrak{m}_B)$ in $F'$ be discrete valuation rings. Then the following are equivalent: 


*

*$\mathfrak{m}_A\subset\mathfrak{m}_B$

*$A\subset B$


Furthermore, we have $A = F\cap B$ and $\mathfrak{m}_A = F\cap \mathfrak{m}_B$.
A prove can be found in Stichtenoth's Algebraic function fields and codes 
Lemma 2 If the local rings of a point on a variety contains the local ring of another point on the same variety, then the points coincide, i.e. $\mathcal{O}_P \subset \mathcal{O}_{P'} \Rightarrow P = P'$.
This is in Hartshorne's Algebraic Geometry I.6.4
Using these facts, we can conclude (using the notation from the question):
Let $R\subset k(C)$ a discrete valuation ring containing $\mathcal{O}_{C',Q}$. Then $R$ corresponds to a point $P$ in $C$. Therefore $\varphi$ maps $P$ to some $Q'$ in $C'$. But the local ring of this $Q'$ is contained in $k(C')\cap R = \mathcal{O}_{C',Q}$, hence $Q' = Q$, and thus $P\in\varphi^{-1}(Q)$.
The only thing, that is not very clear to me now is that I assumed, that this $R$ corresponds to a point $P$ on $C$. For smooth $C$, this would be ok! (Hartshorne I.6.?)
But without this additional assumption, this is not yet clear to me.
